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S Apr 1, 2021 at 23:18 history edited sharpend CC BY-SA 4.0
added comment on tempered dists
S Apr 1, 2021 at 23:18 history suggested Maximilian Janisch CC BY-SA 4.0
added Weyl's Lemma
Apr 1, 2021 at 23:01 review Suggested edits
S Apr 1, 2021 at 23:18
Apr 1, 2021 at 18:11 comment added Maximilian Janisch @AlexandreEremenko Yes absolutely, but $A-B$ can be of the type $A-B: S(\mathbb R^n)\to\mathbb R, \phi\mapsto\int_{\mathbb R^n} p\phi\,\mathrm dx$ for a polynomial $p$. I am pretty sure that that is what Dallas meant 🙂
Apr 1, 2021 at 17:18 comment added Alexandre Eremenko There is no polynomial, except 0, in the space $S(R^n)$, if I correctly understand that $S(R^n)$ is the space of Schwartz test functions.
Apr 1, 2021 at 16:18 vote accept Maximilian Janisch
Apr 1, 2021 at 16:18 comment added Maximilian Janisch @DanieleTampieri Ohhh right, since $A-B$ could be any weak solution to the Laplace equation. Took me a while to see 😆! Thanks to both of you!
Apr 1, 2021 at 16:10 comment added Daniele Tampieri Yes @MaximilianJanisch, you cannot always conclude that $A=B$, and this is in general true for every linear partial differential operator: the non homogeneous equation has a solution which is unique modulo a solution of the homogeneous equation. To make sure that this is only the null one, you must assume additional requirements as sharpened has exemplified.
Apr 1, 2021 at 16:04 comment added sharpend Right, it just isn't true at that level of generality. ($A$ and $B$ could be different constants.) But at least you know what their difference must be. And thank you :)
Apr 1, 2021 at 16:00 comment added Maximilian Janisch So for general $A,B$ we can't always conclude $A=B$ ? 😞 PS: I hope your French is doing well 🙂.
Apr 1, 2021 at 15:57 history answered sharpend CC BY-SA 4.0